Problem 40 - Differential Equations and Linear Algebra

Question

Use the annihilator method to solve the given differential equation. $$y^{\prime \prime}+4 y=7 e^{x}.$$

Step-by-Step Solution

Answer

To solve the given differential equation \(y^{\prime \prime}+4 y=7 e^{x}\) using the annihilator method, we first find the complementary function \(y_c(x) = c_1 \cos{2x} + c_2\sin{2x}\). Next, we find the annihilating equation \(y^{\prime \prime \prime} + 4y^{\prime} - y^{\prime \prime} - 4y=0\) and obtain its general solution \(y(x) = A + c_1\cos{2x} + c_2\sin{2x}\). After applying the initial conditions, we find the particular solution \(y_p(x) = A\). Hence, the overall solution is \(y(x) = A(1 + \cos{2x} +\sin{2x})\).

Step 1: Find the complementary function

To find the complementary function, we need to solve the corresponding homogeneous equation: $$y^{\prime \prime}+4 y=0.$$ This is a linear differential equation with constant coefficients, so we use the characteristic equation to find the solution: $$r^2 + 4 = 0.$$ Solving for r, we get complex roots: $$r = \pm 2i.$$ Thus, our complementary function will be in the form of: $$y_c(x)=c_1 \cos{2x} + c_2\sin{2x}.$$

Step 2: Determine the annihilating differential equation

Our given nonhomogeneous equation was: $$y^{\prime \prime}+4 y=7 e^{x}.$$ To annihilate \(7e^x\), we will integrate it by applying the following annihilator: $$D - 1,$$ where \(D\) is the differentiation operator. So, applying it to \(7e^x\), $$(D - 1)(7e^x) = 7e^x - 7e^x = 0,$$ we get a homogeneous equation. Now apply the same annihilator to the given differential equation: $$(D - 1)(y^{\prime \prime}+4 y - 7e^x) = y^{\prime \prime \prime} + 4y^{\prime} - y^{\prime \prime} - 4y$$

Step 3: Solve the annihilating equation

Now we have the following annihilating equation to solve: $$y^{\prime \prime \prime} + 4y^{\prime} - y^{\prime \prime} - 4y=0.$$ To solve this differential equation, we follow the same procedure used to find the complementary function of the given equation. The characteristic equation is obtained as: $$r^3 + 4r - r^2 - 4 = 0.$$ Factoring out \(r\), we get \((r)(r^2 - r + 4) = 0\). Solving it gives us the roots: \(0, 1 \pm 2i\). Hence, the general solution for annihilating equation is: $$y(x) = A + B\cos{2x} + C\sin{2x}.$$

Step 4: Apply the initial conditions and find general solution

Combining the complementary and particular solutions, we get: $$y(x) = A + c_1 \cos{2x} + c_2\sin{2x}.$$ We can see that the functions corresponding to \(c_1 \cos{2x}\) and \(c_2\sin{2x}\) will be annihilated by the given equation since they are the complementary functions. So, the general solution will be: $$y(x) = A + c_1 \cos{2x} + c_2\sin{2x}.$$

Step 5: Solve for the particular solution

We need to plug the general solution into the given differential equation and find the particular solution. $$y^{\prime \prime}+4 y=7 e^{x}.$$ Plugging in our general solution, we get: $$(-4A + 4c_1) \cos{2x} + (-4A + 4c_2)\sin{2x} = 7 e^{x}.$$ However, this equation is not valid since the left side and right side have different functions. To make it valid, the coefficients of \(\cos{2x}\) and \(\sin{2x}\) on the left side must be equal to zero. Thus, $$4A - 4c_1= 0 \Rightarrow A = c_1,$$ $$4A - 4c_2 = 0 \Rightarrow A = c_2.$$ So, our general solution becomes: $$y(x) = A(1 + \cos{2x} +\sin{2x}).$$ Now, comparing this general solution to the given equation, we can eliminate both \(A\cos{2x}\) and \(A\sin{2x}\) terms as they are complementary parts. Hence, the particular solution is: $$y_p(x) = A.$$

Key Concepts

Complementary Function

The complementary function, denoted as \(y_c(x)\), represents the solution to the corresponding homogeneous equation of a differential equation. In the case of the differential equation \(y^{\text{\textdprime}}+4y=0\), the solution process begins by finding this complementary function. To do this, one solves the characteristic equation \(r^2 + 4 = 0\) which yields complex roots \(r = \text{\textpm} 2i\). Consequently, the complementary function takes the form \(y_c(x) = c_1 \cos{2x} + c_2\sin{2x}\), where \(c_1\) and \(c_2\) are constants determined by initial conditions.

Characteristic Equation

The characteristic equation is a crucial concept in solving linear differential equations with constant coefficients. It is derived from the homogeneous part of a differential equation by substituting \(y = e^{rt}\) and simplifying. For the given problem, when we solve the homogeneous equation \(y^{\text{\textdprime}}+4y=0\), the characteristic equation becomes \(r^2 + 4 = 0\). This equation's roots inform us about the form of the complementary function. Complex roots, such as \(\text{\textpm} 2i\) in our case, lead to the complementary function having sine and cosine terms.

Understanding Roots and Solutions

• Real roots yield exponential functions.
• Complex roots result in trigonometric functions.
• Repeated roots introduce additional polynomial factors.

Particular Solution

The particular solution, \(y_p(x)\), is found by focusing on the nonhomogeneous part of the differential equation. It embodies the 'external' influence represented by the non-zero right side of the differential equation. After obtaining the complementary function for the homogeneous part, we use methods like the annihilator approach to deduce a particular solution. In the sampled problem, the particular solution is determined to be \(y_p(x) = A\), which counts for the nonhomogeneous component when solving \(y^{\text{\textdprime}}+4 y=7 e^{x}\). This represents the influence of the nonhomogeneous term \(7e^x\) on the overall solution.

Homogeneous Equation

A homogeneous equation in the context of differential equations refers to one in which the function and its derivatives are set to zero, such as \(y^{\text{\textdprime}}+4y=0\). This absence of a non-zero term represents a system free from external forces or inputs, allowing us to determine the complementary function \(y_c(x)\). The equation's solutions hold vital properties: linearity and superposition, which imply that the sum of any two solutions is also a solution. The concept of the characteristic equation is tightly linked to the homogeneous part, facilitating the derivation of the complementary function.

Nonhomogeneous Equation

In contrast, the nonhomogeneous differential equation contains a non-zero term that acts as an external force or input, such as the term \(7 e^{x}\) in \(y^{\text{\textdprime}}+4 y=7 e^{x}\). This non-zero term necessitates the determination of a particular solution, \(y_p(x)\), in addition to the complementary function \(y_c(x)\). The annihilator method comes into play when determining a suitable form of the particular solution by 'annihilating' the non-zero term, resulting in an auxiliary homogeneous equation. Solving this new homogeneous equation paves the way to an exhaustive solution that addresses both the homogeneous and nonhomogeneous aspects.